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An alpha nucleus of energy 1/2 mv² bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
Options
(a) 1 / Ze
(b) v²
(c) 1 / m
(d) 1 / v⁴
Correct Answer:
1 / m
Explanation:
Kinetic energy of alpha nucleus is equal to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,
1/2 mv² = 1/4π?￢ﾂﾀ . qαZe / r₀
where r₀ is the distance of closest approach
r₀ = 2/4π?￢ﾂﾀ . qαZe / mv²
r₀ ∞ Ze ∞ qα ∞ 1/m ∞ 1/v²
Hence option (c) is correct.
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two charges each equal to 2µ C are 0.5m apart. If both of them exist inside vacuum
- An electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into
- A coin of mass m and radius r having moment of inertia I about the axis passes
- Two closed pipes produce 10 beats per second when emitting their fundamental nodes.
- Pure Si at 300 k has equal electrons (nₑ) and hole (nh) concentrations
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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